The document discusses expanding and factorizing algebraic expressions. It provides examples of expanding expressions using the distributive property, such as expanding (a + b)(c + d) to get ac + ad + bc + bd. It also discusses factorizing expressions by finding common factors, such as factorizing a2 + 2ab + b2 to get (a + b)2. Tips and techniques are presented for expanding, factorizing, finding common factors, and using the distributive property to manipulate algebraic expressions.

1. CHAPTER 6
Mohd khusaini bin majid
Mrsm kota kinabalu

2. 6.1 CONCEPT OF EXPANDING
BRACKETS
w(x + y) can be expanded to
become wx + wy as
shown below.
Expand a single pair of brackets
w(x + y) = (w x x) + (w x y)
= wx + wy
RECALL
• w x x = wx
• w x y = wy
• wx and wy are unlike
terms.
• unlike terms cannot
be added together

3. Example 1
Expand the following algebraic expressions.
a) r(s – 2t) b) 3e(2e – f + 4g)
a) r(s – 2t) = (r x s) + [r x (-2t)]
b) 3e(2e – f + 4g)
= (3e x 2e) + [3e x (-f)] + (3e x 4g)
= 6e2 – 3ef + 12eg
Solution: TIPS
• (+r) x (+s) = +rs
• (+r) x (-s) = -rs
• (-r) x (+s) = -rs
• (-r) x (-s) = +rs

4. Exercise 6.1A
1. Expand the following algebraic expressions.
a) p(q + r) b) 5(3x + y)
c) –w(x – y) d) -3r(-t +2s)
e) 2q(2r – 3s +t) f) ½p(x + y)
2. Expand the following expressions.
a) -4m(-m – 3n) b) 4g(f/2 – 3g)
c) –n/2(2n + 8m)
d) 4q/3(12p + 6q – 9r)

5. Expand double pairs of brackets
(w + x)(y + z) can be expanded to
wy + wz + xy + xz as shown below.
(w + x)(y + z)
= (w x y) + (w x z) + (x x y) + (x x z)
= wy + wz + xy + xz

6. Example 2
Expand the following algebraic expressions.
a) (2a + b)(c + 3d) b) (m + 4)(m – 3)
a) (2a + b)(c + 3d)
= (2a x c) + (2a x 3d) + (b x c) + (b x 3d)
= 2ac + 6ad + bc + 3bd
Solution:

7. b) (m + 4)(m – 3)
= (m x m) + [m x (-3)] + (4 x m) + [4 x (-3)]
= m2 – 3m + 4m – 12
= m2 + m -12
Like terms
TIPS
• (x x w) = (w x x) = wx
• (y x w) = (w x y) = wy

8. Look at the products of (a + b) and (a – b).
(a + b)(a – b)
= (a x a) + [a x (-b)] + (b x a) + [b x (-b)]
= a2 – ab + ab + b2 = a2 – b2
Since (a + b)(a – b) = (a – b)(a + b),
(a – b)(a – b) is also equal to a2 – b2.
In general, (a + b)(a – b) = (a – b)(a + b)
= a2 – b2

9. Example 3
Expand the following expressions.
a) (a + 2b)(a – 2b) b) (2p + 5)(2p – 5)
a) (a + 2b)(a – 2b)
= (a x a) + [a x (-2b)] + (2b x a) + [2b x (-2b)]
= a2 – 2ab + 2ab – 4b2
= a2 – 4b2 a2 – 4b2 = (a)2 – (2b)2
Solution:

10. b) (2p + 5)(2p – 5)
= (2p x 2p) + [2p x (-5)] + (5 x 2p) + [5 x (-5)]
= 4p2 – 10p + 10p – 25
= 4p2 – 25
Thus, (a + b)(a – b) can be written as a2 – b2
directly.
4p2 – 25 = (2p)2 – (5)2
NOTES
• -10p and 10p are like terms.
• -10p + 10p = 0

11. Example 4
Expand the following expressions.
a) (2x + 7)(2x – 7) b) (3k – 2h)(3k + 2h)
a) (2x + 7)(2x – 7) b) (3k – 2h)(3k + 2h)
= (2x)2 – (7)2 = (3k)2 – (2h)2
= 4x2 – 49 = 9k2 – 4h2
NOTES
(a + b)(a – b) ≠ a2 + b2
Solution:

12. Look at the expansions of (a + b)2 and (a – b)2.
(a +b)2 = (a +b)(a +b)
= (a x a) + (a x b) + (b x a) + (b x b)
= a2 + ab + ba + b2
= a2 + 2ab + b2
TIPS
• ab and ba are like terms.
• ab + ba = 2ab
• -ab – ba = -2ab

13. (a – b)2
= (a – b)(a – b)
= (a x a) + [a x (-b)] + [(-b) x a] + [(-b) x (-b)]
= a2 – ab – ba + b2
= a2 – 2ab + b2
Thus, (a + b)2 = a2 + 2ab + b2
and
(a – b)2 = a2 – 2ab + b2

14. Example 5
Expand the following expressions.
a) (2m + 3n)2 b) (4s – t)2
a) (2m + 3n)2 = (2m)2 + 2(2m)(3n) + (3n)2
= 4m2 + 12mn + 9n2
b) (4s – t)2 = (4s)2 + 2(24s)(t) + (t)2
= 16s2 – 8st + t2
Solution: (a + b)2 = a2 + 2ab + b2
(a - b)2 = a2 - 2ab + b2

15. Exercise 6.1B
1. Expand the following expressions.
a) (a + b)(c + 2d) b) (2m – n)(2p – q)
c) (x + 7)(2y + 5) d) (2p - 7)(q + 2)
e) (u – 2v)(3w + v) f) (3k + 6)(k - 5)
2. Expand the following expressions where each
is the product of the sum and the difference of
two terms.
a) (x + 3y)(x – 3y) b) (f + 5g)(f – 5g)

16. c) (v – 7w)(v + 7w) d) (4h – 3k)(4h + 3k)
e) (5k + 2)(5k - 2) f) (7 + a)(7 – a)
3. Expand the following expressions.
a) (m + n)2 b) (c – d)
c) (3w + 4x)2 d) (5t – u)2
e) (-y + 2)2 f) (-2p - 5q)2

17. 6.2 CONCEPT OF
FACTORIZATION OF
ALGEBRAIC EXPRESSIONS
A number can be
expressed as the product
of its factors.
18 = 1 x 18
= 2 x 9
= 3 x 6
State the factors of an algebraic term
An algebraic term can be
expressed as the product
of its factors.
3ab = 1 x 3ab
= 3 x ab
= a x 3b
= b x 3a

18. State the common factors and the
highest common factor of several
algebraic terms
 Common factors of several Algebraic Terms
Step 1: List all the factors of each algebraic
term.
Step 2: Identify the factors which are common
to all the terms.

19. Example 7
Find the common factors of the algebraic terms,
2ab and 4bc.
2ab = 1 x 2ab
= 2 x ab
= a x 2b
= b x 2a
The factors of 2ab are
1, 2, a, b, 2a, 2b, ab
and 2ab.
Solution:
RECALL
• The common factors are factors
which are common to all the
numbers.
• The HCF is the highest common
factor for all the numbers.
• Examples:
- The factors of 8 are 1, 2, 4 and 8.
- The factors of 12 are 1, 2, 3, 4, 6
and 12.
- The common factors are 1, 2 and 4.
- The HCF is 4.

20. 4bc = 1 x 4bc
= 2 x 2bc
= 4 x bc
= b x 4c
= c x 4b
= 2b x 2c
The factors of 4bc are 1, 2, 4, b, c, 2b, 2c, 4b,
4c, bc, 2bc and 4bc.
Thus, the common factors of 2ab and 4bc are
1, 2, b and 2b.

21. Example 8
 Highest Common Factor (HCF) of Several
Algebraic Terms
Find the highest common factor of the algebraic
terms, 3uv and 6u2.
3uv = 1 x 3uv
= 3 x uv
= u x 3v
= v x 3u
Solution:
The factors of 3uv are
1, 3, u, v, 3u, 3v, uv
and 3uv.

22. 6u2 = 1 x 6u2
= 2 x 3u2
= 3 x 2u2
= 6 x u2
= u x 6u
= 2u x 3u
The common factors of 3uv and 6u2 are 1, 3, u
and 3u.
Thus, the highest common factor of 3uv and
6u2 is 3u.
The factors of 6u2 are
1, 2, 3, 6, u, 2u, 3u, 6u,
u2, 2u2, 3u2 and 6u2.

23. ANOTHER WAY
Using the division method to find the HCF.
 Divide the terms by a common factor until no more common
factor exists.
The HCF of 3uv and 6u2 = 3 x u
= 3u
This is a faster way of determining HCF where many algebraic
terms are involved.
3uv, 6u2
uv, 2u2
v, 2u
3
u

24. Exercise 6.2A
Find the highest common factor (HCF) of the
following algebraic terms.
1. 2f, 4g 2. 4xy, 8yz
3. 5g2, 10gh 4. n2p, np2
5. 8km2, 12mn 6. ef2, 2ef, 4efg
7. 6abc2, 12a2bc, 18ab2c

25. Factorize algebraic expressions
Factorization is the reverse process of expansion.
NOTES
Expansion
p(q + r) = pq + pr
Factorization

26. Factorize ab – ac
Consider an expression 2ab – 4ac.
2ab, 4ac
ab, 2ac
b, 2c
2
a
HCF = 2 x a
= 2a
NOTES
2ab – 4ac = 2ab – 4ac
= b – 2c
2a 2a2a
2ab – 4ac = 2a(b – 2c)
HCF Final
quotient
Thus, 2ab – 4ac can be
expressed as the product of
2a and (b – 2c).

27. Example 9
Factorize the following algebraic expressions.
a) 2pq + 4qr b) mn – mn2
c) 3xy + 6yz – 9y2
Solution:
a) 2pq + 4qr = 2q(p + 2r)
HCF Final quotient
2pq, 4qr
pq, 2qr
p, 2r
2
q
HCF = 2q
Check:
2q(p + 2r) = 2pq + 4qr

28. b) mn – mn2 = mn(1 – n)
c) 3xy + 6yz – 9y2 = 3y(x + 2z – 3y)
mn, mn2
n, n2
1, nHCF Final quotient
3xy, 6yz, 9y2
xy, 2yz, 3y2
x, 2z, 3y
3
y
m
n
HCF = 3y
HCF = mn
HCF Final quotient
Check:
3y(x + 2z – 3y) = 3xy + 6yz – 9y2
Check:
mn(1 - n) = mn – mn2

29. Exercise 6.2B
Factorize the following algebraic expressions.
a) 2x + 8y b) 6d + 12
c) 6uv – 3v d) 6m2 – 4mn
e) 10x2y2 – 5xy f) 36fg + 27g2
g) 3a – 6b + 9c h) u2 – uv + uw

30. Factorize a2 + 2ab + b2
To factorize a2 + 2ab + b2, we can write 2ab as
ab + ab.
Hence, a2 + 2ab + b2 = a2 + ab + ab + b2
= a(a + b) + b(a + b)
= (a + b)(a + b)
= (a + b)2
Thus, a2 + 2ab + b2 = (a + b)(a + b)
= (a + b)2
(a + b) is the
common factor.
The HCF of a2 and ab is a. Thus, a2 + ab = a (a + b).
The HCF of ab and b2 is b. Thus, ab + b2 = b(a + b).

31. Example 10
Factorize the following algebraic expressions.
a) x2 + 8x + 16 b) y2 + 6yz + 9z2
a) x2 + 8x + 16 = x2 + 4x + 4x + 16
= x(x + 4) + 4(x + 4)
= (x + 4)(x + 4)
= (x + 4)2
Solution:
The HCF of x2 and 4x is x. So x2 + 4x = x (x + 4).
The HCF of 4x and 16 is 4. So 4x + 16 = 4(x + 4).
Check:
(x + 4)2 = (x)2 + 2(x)(4) + (4)2
= x2 + 8x + 16
TIPS
• 8x = 4x + 4x
• 6yz = 3yz + 3yz

32. b) y2 + 6yz + 9z2 = y2 + 3yz + 3yz + 9z2
= y(y + 3z) + 3z(y + 3z)
= (y + 3z)(y + 3z)
= (y + 3z)2
Check:
(y + 3z)2 = (y)2 + 2(y)(3z) + (3z)2
= y2 + 6yz + 9z2

33. Exercise 6.2C
Factorize the following algebraic expressions.
1. k2 + 8k + 16 2. h2 + 6h + 9
3. b2 + 12b + 36 4. u2 + 10uv + 25v2
5. 16y2 + 8yz + z2 6. 4t2 + 12t + 9

34. Example 11
Factorize the following algebraic expressions.
a) ax + 3a + 2x + 6
b) ab + by + ax + xy
Factorize ab + ac + bd + cd
TIPS
• Group two terms in the expressions as one
group.
• Find the HCF of each group.
• Divide each group by the HCF to get the
final quotient.
• Make sure the two final quotients are the
common factors.

35. a) ax + 3a + 2x + 6
= (ax + 3a) + (2x + 6)
= a(x + 3) + 2(x + 3)
= (x + 3)(a + 2)
b) ab + by + ax + xy
= (ab + by) + (ax + xy)
= b(a + y) + x(a + y)
= (a + y)(b + x)
Solution:
Common factor
Find the HCF of
the expressions.
Common factor
Find the HCF of
the expressions.

36. Exercise 6.2D
Factorize the following algebraic expressions.
1. 4ac + 2ab + 2cd + bd
2. st + 4tv + 3su + 12uv
3. 2ab + 2ac + b + c
4. 4x2 - 2xy - 6xz + 3yz

37. Factorize a2 – b2
Since factorization is the reverse process of
expansion, a2 – b2 can be factorized as
(a + b)(a – b).
a2 – b2 is also known as the difference of two
squares.
a2 – b2 = (a + b)(a - b)

38. Example 12
Factorize the following algebraic expressions.
a) 4x2 – y2 b) 8p2 – 18
a) 4x2 – y2 b) 8p2 - 18
= 22x2 – y2 = 2(4p2 – 9)
= (2x)2 – y2 = 2[(2p)2 – 32]
= (2x + y)(2x - y) = 2(2p + 3)(2p – 3)
Solution:

39. Exercise 6.2E
Factorize the following algebraic expressions.
1. p2 – q2 2. a2 - 32
3. x2 – 9 4. 1 – 9t2
5. 16p2 – 25q2 6. 81u2 – 100
7. 2x2 – 18y2 8. 18gh2 – 50g

40. Factorize and simplify algebraic
fractions
An algebraic fraction is one in which the
numerator or the denominator or both are
algebraic expressions. Examples:
3a2 , 8 , 12gh2k , 9p + 6pq , x2 – y2
6 4ef 3gh 6 + 4q x2 + xy

41. Example 13
Simplify the following expressions.
a) 4ab2 b) 6pq
a) 4ab2 = 2a(2b)
= 2b2
6a2c 9p2qr
2a(3ac)6a2c
3ac
• The common factor of 4ab2 and 6a2c is 2a.
• Divide both numerator and denominator by 2a.
No more common factor
Solution:

42. b) 6pq = 3pq(2)
= 2
9p2qr 3pq(3pr)
3pr
• The common factor of 6pq and 9p2qr is
3pq.
• Divide both numerator and denominator
by 3pq.
No more common factor

43. Example 14
Simplify the following expressions.
a) 2x + 10 b) m2 – 9
a) 2x + 10
= 2(x + 5)
= 2
xy + 5y mn – 3n
xy + 5y
y(x + 5)
y
• Factorize both numerator and denominator.
• Divide by the common factor (x + 5).
Solution:

44. b) m2 – 9
= (m + 3)(m - 3)
= m + 3
mn – 3n
n(m – 3)
n
• Factorize both numerator and
denominator.
• Divide by the common factor (m - 3).

45. Exercise 6.2F
1. Simplify the following algebraic fractions.
a) 5ab b) 6p
c) 9pq d) fg
e) w2u f) 12b2c2
n
10bc 18pr
wu
4fgh
abc

46. 2. Simplify the following algebraic fractions to
the lowest terms.
a) 12m + 3 b) 4a2 – 12ab
c) pq , d) u + v
e) s – t f) f2 – 9
f2 – 3fs2 – t2
5u + 5vpr - ps
6ab6p

47. 6.3 ADDITION AND
SUBTRACTION OF
ALGEBRAIC FRACTIONS
Add or subtract two
algebraic fractions
with the same
denominator
TIPS
• Add or subtract the numerators.
• Retain the denominators.
• + (a + b) = +a + b
+ (a – b) = +a – b
– (a + b) = - a – b
– (a – b) = - a + b

48. Example 15
Simplify the following.
a) m + 3 + 2m + 1 b) k - k – 1
4 h + k4 h + k
a) m + 3 + 2m + 1 = (m + 3) + (2m + 1)
= m + 3 + 2m + 1
= 3m + 4
4 4 4
4
4
Solution:

49. b) k – k - 1 = k – (k – 1)
= k – k + 1
= 1
h + kh + k h + k
h + k
h + k

50. Exercise 6.3A
Simplify the following to the lowest terms.
1. a + b 2. 9x – 2x
3. mn + 3mn 4. 10 + 3
5. 3a + 1 + a + 2 6. 7w - 4w
7. 2p + 5 + p - 2 8. b + 1 + b + 1
3 3 11 11
8 8 d2d2
4b2 4b2 15wuv15wuv
3q 3q 2b + 12b + 1

51. Add or subtract two algebraic
fractions with one denominator as a
multiple of the other denominator
Simplify the following to the lowest terms.
a)
Example 16

52. Exercise 6.3B
Simplify the following to the lowest terms.
1. a + a 2. 2p – 4p
3. y + y + 2 4. b + d
5. 2k + 4k 6. f + g – g - f
3 9 5 15
2 6 rstst
3(m + 2)m + 2 2hkhk

53. Exercise 6.3C
1. Simplify each of the following.
a) a + a b) h - 3
c) b + a d) 1 - b
e) 3y + 5q f) 3d – 2a
3da7
ac
4b
2bb2a
nm52

54. 2. Simplify each of the following algebraic
fractions.
a) x + y b) p - q
c) h – h d) a + b
e) b + d - c f) w + w + u
96
2k k
86
strt
babpqp

55. 6.4 MULTIPLICATION AND
DIVISION OF ALGEBRAIC
FRACTIONS
a x c = a x c
= ac
Multiplication of two algebraic
fractions
b d b x d
bd

56. Example 19

57. Exercise 6.4A
Find the product of the following.
1. p x 5q 2. (w + u) x x
3. 2t x w 4. 5 x (-h)
5. (x + y) x z 6. 4 x p + q
3 y
v2 u2 6g
x2 + 1 p - q 5

58. Exercise 6.4B
Find the quotient of the following.
1. m ÷ 5n 2. m ÷ n
3. b2c ÷ a 4. 5 ÷ (-u)
5. (x + y) ÷ z + 1 6. s ÷ a ,
3 p qr
d 3c 6w
w p - q b + d

59. Exercise 6.4C
1. Find the product of the following.
a) p x 3pq b) 15xy x 4
c) uv x x2 d) p x (-qr2)
e) – rs2 x uv f) 4 x p2 – q2
q 20x
wx u2 qr
4tu r2s p - q 5

60. 2. Divide the following.
a) m ÷ mn b) xyz ÷ wx
c) u2 ÷ (-u) d) (ax + ay) ÷ x + y
e) s ÷ a f) a + b ÷ a2 – b2
3 y
6w w
p2 – q2 p - q c - d 3(c – d)

61. SUMMARY
Expansion
 The product of an algebraic
term and an algebraic
expression a(b + c) = ab + ac
 The product of two algebraic
expressions (a + b)(c + d)
= ac + ad + bc + cd
(a + b)2 = a2 + 2ab + b2
(a – b)2 = a2 – 2ab + b2
(a + b)(a – b) = a2 – b2
ALGEBRAIC
EXPRESSIONS III
Factorization
 Factorize an algebraic term 2ab =
2 x a x b
 Factorize algebraic expressions
ab + ac = a(b + c)
a2 - b2 = (a + b)(a – b)
a2 + 2ab + b2 = (a + b)2
a2 – 2ab + b2 = (a – b)2
ab + ac + bd + cd = (b + c)(a + d)
Algebraic fractions
Both or either the numerator or the denominator is an algebraic
term or algebraic expression.
b , 3 , a + b , b , and c – d
a + baa5 a + b